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erastovalidia [21]
2 years ago
14

What is the sum of the coefficients for the balanced chemical equation below ?

Chemistry
1 answer:
choli [55]2 years ago
6 0

Answer:

9

4X \:  +  \: 3Y _{2} \: → \: 2X _{2} Y _{3}

Sum = 4+3+2

=9

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If 190dm of hydrogen gas collected at 20°c and 760mmHg .Calculate it's volume at stp (standard pressure=760mmHg
Andrei [34K]

Answer:

177.1 L

Explanation:

The excersise can be solved, by the Ideal Gases Law.

P . V = n . R . T

In first step we need to determine the moles of gas:

We convert T° from, C° to K → 20°C + 273 = 293K

We convert P from mmHg to atm → 760 mmHg = 1atm

1Dm³ = 1L → 190L

We replace: 190 L . 1 atm = n . 0.082 . 293K

(190L.atm) / 0.082 . 293K = 7.91 moles.

We replace equation at STP conditions (1 atm and 273K)

V = (n . R .T) / P

V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L

We can also make a rule of three:

At STP conditions 1 mol of gas occupies 22.4L

Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L

3 0
3 years ago
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JulsSmile [24]

Answer:

a metal spoon left in boiling water

Explanation:

5 0
3 years ago
The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w
Alexus [3.1K]

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

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The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

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